I need some help solving this. I have tried:
$$ \begin{bmatrix} a & b \\ c & d \\ \end{bmatrix} =\frac{1}{\operatorname{det}A}\cdot \begin{bmatrix} d & -b \\ -c & a \\ \end{bmatrix}$$ I ended up with $$a=\frac{d}{\operatorname{det}A},$$ and $$d=\frac{a}{\operatorname{det}A}.$$ Then $$\operatorname{tr}(A)=a+d=\frac{a+d}{\operatorname{det}A},$$ but I don't really think it works.
On
You have $A^{-1} = A \implies A^2 = I$. So we just calculate that for the matrix you have $$\begin{pmatrix} 1 & 0 \\ 0& 1 \end{pmatrix} = \begin{pmatrix} a & b \\ c& d \end{pmatrix}\cdot\begin{pmatrix} a & b \\ c& d \end{pmatrix} = \begin{pmatrix} a^2 + bc & (a+d)\cdot b \\ (a+d)\cdot c & bc + d^2 \end{pmatrix}$$
Therefore, you get $(a+d)\cdot b = (a+d)\cdot c = 0$.
Case 1: $a+d = 0$, we're done.
Case 2: Let's assume $b$ or $c$ are $0$ then $a^2 = d^2 = 1 \implies a^2 - d^2 = 0 \implies a = \pm d$.
If $a = -d \implies a+d = 0$
If $a = d$, then $A = aI$, and $A^2 = I \implies a = \pm 1 \implies A = \pm I$
Therefore, for $A^2 = I$, we have $\text{tr} A = 0, 2 , -2$
Continuing from your chain of thought
In your calculations, you obtained the following,
$$a+d = \frac{a+d}{\det A}$$ Now, as we know if $A^2 = I$, then $\det A = \pm1$.
Therefore, for $\det A = -1$. You get $$a+d = -(a+d) \implies \text{tr} A = 0$$
For $\det A = 1$, you get $$\begin{pmatrix} a & b\\ c & d \end{pmatrix} = \begin{pmatrix} d & -b\\ -c & a \end{pmatrix}$$
$\implies b = c = 0, a = d$. Therefore $A = aI$, and $\text{tr} A = 2a$ $$A^2 = I \implies (aI)^2 = I \implies a^2 = 1 \implies a = \pm 1$$
Hence, $\text{tr} A = \pm 2$
Comment: For a more 'clean' answer, please refer to Omnomnomnom's answer/response. I provided an elementary answer to demonstrate one exists, but the better and faster way, in my opinion, is still using eigenvalues.
On
By the Cayley–Hamilton theorem or direct verification, we have $A^{2}-\operatorname {tr}(A)A+\det(A)I=0$.
From $A^2=I$, we get $\operatorname {tr}(A)A=(\det(A)+1)I$.
Taking traces on both sides, we get $\operatorname {tr}(A)^2=2(\det(A)+1)$.
From $A^2=I$, we also get $\det(A)^2=1$ and so $\det(A)=\pm1$.
If $\det(A)=1$, then $\operatorname {tr}(A)^2=4$ and so $\operatorname {tr}(A)=\pm 2$.
If $\det(A)=-1$, then $\operatorname {tr}(A)^2=0$ and so $\operatorname {tr}(A)=0$.
These three possibilities occur for the matrices below: $$ \operatorname {tr}\begin{pmatrix}1&0\\0&1\end{pmatrix} = 2 \qquad \operatorname {tr}\begin{pmatrix}-1&\hphantom-0\\\hphantom-0&-1\end{pmatrix} = -2 \qquad \operatorname {tr}\begin{pmatrix}1&\hphantom-0\\0&-1\end{pmatrix} = 0 $$
This problem is easier if you think about it in terms of the eigenvalues of $A$. Note that if $x$ is an eigenvector of $A$, we have $A^{-1}x = Ax$. What does this tell you about that eigenvalue?